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 January 29th, 2013, 06:09 AM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 prove inequality $a,b,c \in R$ satisfying :a+b+c=0 and , abc=8 prove : $\frac {1}{a}+\frac {1}{b}+\frac {1}{c}<0$
 January 29th, 2013, 06:13 AM #2 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: prove inequality Without loss of generality, let a <= b <= c. We must have a < b < 0 and so |a| < |c| and so 1/a + 1/c < 0 and hence 1/a + 1/b + 1/c < 0.
 January 29th, 2013, 06:19 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: prove inequality $\frac{1}{a}\,+\,\frac{1}{b}\,+\,\frac{1}{c}\,=\,\f rac{bc\,+\,ac\,+\,ab}{8}\,=\,\frac{(a\,+\,b\,+\,c) ^2\,-\,a^2\,-\,b^2\,-\,c^2}{16}\,<\,0=$
 January 29th, 2013, 07:48 AM #4 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: prove inequality greg1313: very good solution

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